I was trying to bound the value of $v_p(2^n-1)$ and some of the series I obtain made me wonder about the following problem.
Problem : When does the series $$\sum_{prime \: p} \frac{1}{(ord_p 2)^z} = \sum_{n=1}^{\infty} \frac{\#\{p|ord_p2 = n\}}{n^z}$$ converge if $ord _p 2$ denotes the multiplicative order of $2$ in $\mathbb{Z}/p\mathbb{Z}$ for an odd prime $p$?
I noticed that it diverges when $z=1$ since $\#\{p|ord_p 2 = n\} \ge 1$ for all $n \ge 2$ except for $n=6$. Also, it converges when $z > 2$ because $$\sum_{n=1}^{\infty} \frac{\#\{p|ord_p2 = n\}}{n^z} \le \sum_{n=1}^{\infty} \frac{\sum_{ord_p2=n} \log p}{n^z} \le \sum_{n=1}^{\infty} \frac{\log (2^n -1)}{n^z}$$ I'm especially interested about what happens when $z=2$.
Thanks in advance!